# Applied optimization

Literally, optimization is making something the best, but we use it in math to mean maximization, which is making something the biggest. (You can imagine that the thing that you're maximizing is a numerical measure of how good the thing that you're optimizing is.) Essentially the same principles apply to minimization, which is making something the smallest. (And pessimization is making something the worst, although people don't use that term very much.) A generic term for making something the largest or smallest is extremization.

The key principle of optimization is this:

A quantity u can only take a maximum or minimum value when its differential du is zero or undefined.
If you write u as f(x), where f is a fixed differentiable function and x is a quantity whose range of possible values you already understand (typically an interval), then du = f'(x) dx. So u can only take an extreme value when its derivative (with respect to x) is zero or undefined or when you can no longer vary x smoothly however you please (which must occur at the extreme values of x and typically only then). This recreates the situation that we've discussed before, finding the extreme values of a function defined on an interval. However, the principle that du is zero or undefined applies even when u is not explicitly given as a function of anything else.

Be careful, because u might not have a maximum or minimum value! Assuming that u varies continuously (which it must if Calclulus is to be useful at all), then it must have a maximum and minimum value whenever the range of possibilities is compact; this means that if you pass continuously through the possibilities in any way, then you are always approaching some limiting possibility. (In terms of u = f(x), this is the case when f is continuous and its domain, the range of possible values of x, is a closed and bounded interval.) However, if the range of possibilities heads off to infinity in some way, or if there is an edge case that's not quite possible to reach, then you also have to take a limit to see what value u is approaching. If any such limit is larger than every value that u actually reaches (which includes the possibility that a limit is ∞), then u has no maximum value; if any such limit is smaller than every value that u actually reaches (which includes the possibility that a limit is −∞), then u has no minimum value.

So in the end, you look at these possibilities:

• when the derivative of u is zero or undefined,
• the extreme edge cases, and
• the limits approaching impossible limiting cases.
Whichever of these has the largest value of u gives you the maximum, and whichever has the smallest value of u gives you the minimum; but if the largest or smallest value is only approached in the limit, then the maximum or minimum technically does not exist.

Here is a typical problem: The hypotenuse of a right triangle (maybe it's a ladder leaning against a wall) is fixed at 20 feet, but the other two sides of the triangle could be anything. Still, since it's a right triangle, we know that x2 + y2 = 202, where x and y are the lengths of legs of the triangle. Differentiating this, 2x dx + 2y dy = 0. Now suppose that we want to maximize or minimize the area of this triangle. Since it's a right triangle, the area is A = ½xy, so dA = ½y dx + ½x dy. If this is zero, then ½y dx + ½x dy = 0, to go along with the other equation 2x dx + 2y dy = 0.

The equations at this point will always be linear in the differentials, so think of this is a system of linear equations in the variables dx and dy. There are various methods for solving systems of linear equations; I'll use the method of addition aka elimination, but any other method should work just as well. So ½y dx + ½x dy = 0 becomes 2xy dx + 2x2 dy = 0 (multiplying both sides by 4x), while 2x dx + 2y dy = 0 becomes 2xy dx + 2y2 dy = 0 (multiplying both sides by y). Subtracting these equations gives (2x2 − 2y2) dy = 0, so either dy = 0 or x2 = y2. Now, x and y can change freely as long as they're positive, but we have limiting cases: x → 0+ and y → 0+. Since x2 + y2 = 400, we see that x2 → 400, so x → 20, as y → 0. Similarly, y → 20 as x → 0. In those cases, A = ½xy → 0. On the other hand, if x2 = y2, then x = y, so x, y = 10√2, since x2 + y2 = 400. In that case, A = ½xy = 100.

So the largest area is 100 square feet, and while there is no smallest area, the area can get arbitrarily small with a limit of 0.

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This web page was written in 2015 by Toby Bartels, last edited on 2015 November 17. Toby reserves no legal rights to it.

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